!62
Member
- Location
- metro Atlanta, GA
Hello;
I have a 480vac three phase combination series parallel heating element (resistive load) totaling 2779 watts in a delta configured circuit. After a nodal analysis and KCL, my REQ ckt. is three elements in a delta configuration with two elements of 213 ohms and one element of 371 ohms with a total of approx. 3.3 amps drawn, and I used a 4 amp breaker. I know that only 480v is applied between any two lines at a time which is single phase or the √2; however, I used the two following equations to determine Line Phase Current:
Ln1 - Ln2 = 1079w and 213Ω (due to series parallel elements)
Ln2 - Ln3 = 1079w and
#2). Ln1 = √A2 + C2 + (A x C),
Ln2 = √B2 + A2 + (B x A), and
Ln3 = √C2 + B2 + (C x B).
Can anyone confirm these equations and/or give an equation to complete the circuit.
As well, I did use 2779w/480v x √3 = 3.34A total draw; this would be per phase.
I have a 480vac three phase combination series parallel heating element (resistive load) totaling 2779 watts in a delta configured circuit. After a nodal analysis and KCL, my REQ ckt. is three elements in a delta configuration with two elements of 213 ohms and one element of 371 ohms with a total of approx. 3.3 amps drawn, and I used a 4 amp breaker. I know that only 480v is applied between any two lines at a time which is single phase or the √2; however, I used the two following equations to determine Line Phase Current:
Ln1 - Ln2 = 1079w and 213Ω (due to series parallel elements)
Ln2 - Ln3 = 1079w and
213Ω
Ln3 - Ln1 = 621w and 371(due to series parallel elements)
Ω (due to series element)
#1). ILine = 1.8A x √3, or#2). Ln1 = √A2 + C2 + (A x C),
Ln2 = √B2 + A2 + (B x A), and
Ln3 = √C2 + B2 + (C x B).
Can anyone confirm these equations and/or give an equation to complete the circuit.
As well, I did use 2779w/480v x √3 = 3.34A total draw; this would be per phase.
